为了好玩，只需要一行语句。

(lambda iterable: reduce(lambda (uniq, dup), item: (uniq, dup | {item}) if item in uniq else (uniq | {item}, dup), iterable, (set(), set())))(some_iterable)

我会用熊猫做这个，因为我经常用熊猫

import pandas as pd
a = [1,2,3,3,3,4,5,6,6,7]
vc = pd.Series(a).value_counts()
vc[vc > 1].index.tolist()

给了

[3,6]

可能不是很有效，但它肯定比许多其他答案的代码更少，所以我想我可以贡献一下

为了实现这个问题，我们可以使用多种不同的方法来解决它，这两种是常见的解决方案，但在实际场景中实现它们时，我们还必须考虑时间复杂性。

import random
import time

dupl_list = [random.randint(1,1000) for x in range(500)]
print("List with duplicate integers")
print (dupl_list)


#Method 1 
print("******************Method 1 *************")

def Repeat_num(x):
    _size = len(x)
    repeated = []
    for i in range(_size):
        # print(i)
        k = i + 1
        for j in range(k, _size):
            # print(j)
            if x[i] == x[j] and x[i] not in repeated:
                repeated.append(x[i])
    return repeated

start = time.time()
print(Repeat_num(dupl_list))
end = time.time()
print("The time of execution of above program is :",(end-start) * 10**3, "ms")

print("***************Method 2****************")

#method 2 - using count()
def repeast_count(dup_list):
  new = []
  for a in dup_list:
      # print(a)
      # checking the occurrence of elements
      n = dup_list.count(a)
      # if the occurrence is more than
      # one we add it to the output list
      if n > 1:
          if new.count(a) == 0:  # condition to check
              new.append(a)
  return new


start = time.time()
print(repeast_count(dupl_list))
end = time.time()
print("The time of execution of above program is :",(end-start) * 10**3, "ms")

# #输出示例:

List with duplicate integers
[5, 45, 28, 81, 32, 98, 8, 83, 47, 95, 41, 49, 4, 1, 85, 26, 38, 82, 54, 11]
******************Method 1 *************
[]
The time of execution of above program is : 1.1069774627685547 ms
***************Method 2****************
[]
The time of execution of above program is : 0.1881122589111328 ms

对于一般的理解，方法1是好的，但是对于真正的实现，我更喜欢方法2，因为它比方法1花费的时间更少。

我注意到大多数解决方案的复杂度为O(n * n)，对于大型列表来说非常缓慢。所以我想分享一下我写的函数，它支持整数或字符串，在最好的情况下是O(n)。对于一个包含10万个元素的列表，最上面的解决方案需要超过30秒，而我的解决方案只需0.12秒

def get_duplicates(list1):
    '''Return all duplicates given a list. O(n) complexity for best case scenario.
    input: [1, 1, 1, 2, 3, 4, 4]
    output: [1, 1, 4]
    '''
    dic = {}
    for el in list1:
        try:
            dic[el] += 1
        except:
            dic[el] = 1
    dupes = []
    for key in dic.keys():
        for i in range(dic[key] - 1):
            dupes.append(key)
    return dupes


list1 = [1, 1, 1, 2, 3, 4, 4]
> print(get_duplicates(list1))
[1, 1, 4]

或者获得唯一的副本:

> print(list(set(get_duplicates(list1))))
[1, 4]

第三个接受答案的例子给出了一个错误的答案，并且没有试图给出重复的答案。下面是正确的版本:

number_lst = [1, 1, 2, 3, 5, ...]

seen_set = set()
duplicate_set = set(x for x in number_lst if x in seen_set or seen_set.add(x))
unique_set = seen_set - duplicate_set

some_list = ['a', 'b', 'c', 'b', 'd', 'm', 'n', 'n']
some_dictionary = {}

for element in some_list:
    if element not in some_dictionary:
       some_dictionary[element] = 1
    else:
        some_dictionary[element] += 1

for key, value in some_dictionary.items():
    if value > 1:
       print(key, end = ' ')

# another way
duplicates = []

for x in some_list:
    if some_list.count(x) > 1 and x not in duplicates:
        duplicates.append(x)

print()
print(duplicates)

来源:这里